When 5.00 mol of benzene is vaporized at a constant pressure of 1.00 atm and at its normal boiling point of 80.1°C,169.5 kJ are absorbed and PΔV for the vaporization process is equal to 14.5 kJ then
A) ΔE = 155.0 kJ and ΔH = 169.5 kJ.
B) ΔE = 184.0 kJ and ΔH = 169.5 kJ.
C) ΔE = 169.5 kJ and ΔH = 184.0 kJ.
D) ΔE = 169.5 kJ and ΔH = 155.0 kJ.
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